of of Systems with Internal Angular -II ManuelM. Dorado Dorado∗ CiRTA,ENITA, SL. S.A., Pol. Ind. Miguel Tres Yuste, Cantos12, Oeste, 28037 s/n Madrid, Spain Abstract Using the Euler’s equations and the Hamiltonian formulation, an attempt has been made to obtain the of systems with internal that are moving with respect to a reference frame when subjected to an interaction. This interaction involves the application of a that is permanently to the internal angular momentum vector.

) L , whose centre of mass moves at a constant ν with respect to a reference frame which INTRODUCTION canbedefinedassoonastheinteractioninthe starts. It is aimed to obtain the equations The angular momentum of a many- sys- of motion to describe the dynamical behaviour of tem with respect to its centre of mass is known as the system from the instant that it undergoes an M the ”internal angular momentum” and is a prop- interaction, by applying a torque that is perma- erty of the system that is independent of the ob- nently perpendicular to the internal angular mo- server. Internal angular momentum is therefore mentum (See Fig.1). and attribute that characterizes a system in the When solving the problem, the following points same way as its mass or charge. In the case of a are taken into account: and particularly in the case of an ele- (a) The cylinder will continue spinning about its mentary particle, the internal angular momentum longitudinal axis at a constant angular veloc- is also referred to as ””. ity ω throughout all the movement. In other To the author’s knowledge, the dynamical be- words, its internal angular momentum mod- haviour of systems with internal angular momen- ule is constant. The that the cylinder tum has not been systematically studied within the possesses as a result of its about its framework of classical (See References longitudinal axis is consequently considered to [1]). This paper approaches this type of problem be internal and does not interfere with its dy- through a model comprising a rotating cylinder, namical behaviour. with constant velocity of rotation, around its longi- tudinal axis. The equations of motion are obtained (b) The of the internal angular momen- using the Euler’s equations and by the Hamiltonian tum, L , with respect to a frame of axes of iner- procedure. tial reference (X, Y , Z) satisfies the Results coincide when the problem is solved us-     ing vectorial algebra or Lagrangian formalism (un- dL dL × L published). However, as a result of changes in ”per- dt = dt + Ω (1) spective”, each method uncovers new peculiarities XY Z XY Z regarding the intimate nature of the system’s be- haviour. in which Ω is the rotation velocity of the frame linked to the solid (X, Y , Z)abouttheframe X Y Z FORMULATION OF THE PROBLEM of inertial reference axes ( , , ). (c) Only infinitesimal are considered that Let us consider a rigid cylindrical solid with internal are compatible with the system’s configura- angular momentum (with respect to its centre of tional variations brought about by the applied ∗ Email:E-mail: [email protected] [email protected] torque. (d) The total energy of the system is not explicitly (a) ω: the rotational velocity of the body around dependent on . its longest axis (let us call it the intrinsic ro- tational velocity of the body). (e) No term is included that refers to the potential energy. According to hypothesis, the applied (b) Ω: the rotational velocity of the system of co- X Y torque is a null acting on the system and ordinates associated with the body ( , , Z X Y the possibility of including a potential from ) viewed from the frame of ( , , Z which the applied torque derives is unknown. ). We can write L and M , refered to (X, Y , Z), as follows:

L = I1ω1e1 + I2ω2e2 + I3ω3e3 (3) Ω=Ω1e1 +Ω2e2 +Ω3e3 (4) M = M1e1 + M2e2 + M3e3 (5)

So, the equation (2) takes the expression

I1ω˙ 1 + I3ω3Ω2 − I2ω2Ω3 = M1 I2ω˙ 2 + I1ω1Ω3 − I3ω3Ω1 = M2 (6) I3ω˙ 3 + I2ω2Ω1 − I1ω1Ω2 = M3 Figure 1. Formulation of the problem These equations are known as modified Euler’s equations and can also be found in (1.e). HOW TO APPROACH THESE PROB- In the case treated above, we must substitute the LEMS THROUGH THE EULER’S EQUA- angular momentum TIONS? L = I3ω3e3 (7) The figure 1 shows a cylindrical rigid body with in (6): angular momentum, L , about its longest axis. We should consider two referential frames: one associ- I3ω3Ω2 = M1 ated with the solid (X, Y , Z), having the Z axis I3ω3Ω1 = M2 (8) on the direction of the body’s longest axis. We will I ω M refer this system of coordinates to a inertial frame 3 ˙ 3 = 3 (X, Y , Z). And then we get In a given instant, a torque M is applied on the M1 rigid body. Ω2 = We call Ω to the rotational velocity of the sys- I3ω3 tem of coordinates associated with the solid (X, M2 Ω1 = (9) Y , Z ), viewed from the inertial frame (X, Y , Z). I3ω3 From here, we will consider the system of coordi- M3 ω˙ 3 = nates (X , Y , Z ) all along the . I3 Assuming this situation, it is known that If the torque is on the axis Y , M1 =0,M3 =0     dL dL and: = + Ω × L (2) dt dt 2 XY Z XY Z Ω =0 M2 Ω1 = (10) It is clear that we have two different I3ω3 acting: ω˙ 3 =0 Where Ω1 coincides with the known velocity of Looking at the results arising from the previous . study, we can say that the body rotates following We have obtained that the body moves along a a circular , with a given by the last circular with a rotational velocity Ω1, expression. This circular trajectory implies that but this is not enough to determinate the radius of the velocity vector must precess jointly with the the orbit. intrinsic angular momentum vector. From this last The principle of conservation for the energy conclusion it can be proved that the force needed should lead us to the expression for this radius. to cause the particle to draw a circular trajectory Assuming our initial hypothesis, the are is expressed as: applied perpendicular to the of movement of F mν × the solid and so, these forces do not produce any = Ω (15) work while the cylinder is moving. When the internal angular of the system We have two conditions that have to be satisfied does not coincide with one of the principal axes of by the body while rotating: inertia, the treatment of the problem is much more M complex, but as it will be seen in the next section, (a) Ω = L the results given above completely general. (b) The total energy should remain constant. HAMILTONIAN FORMULATION. Before the torque is applied the energy of the EQUATIONS OF MOTION AND THEIR cylinder takes the following expression: SOLUTION

1 2 1 2 E = mνo + Iω0 (11) The equations of motion are obtained by the Hamil- 2 2 tonian formulation, where the independent vari- When the torque is acting on the cylinder about ables are the and moments. q q t the defined by the e2, it rotates To do this, a change of the frame ( ,˙, )to p q t along a trajectory, generally defined by r(t)and ( , , ) is made using the Legendre transformation. θ˙(t) that, in this case, coincides with Ω(t)andits From the  energy takes the following expression: H(p, q, t)= q˙ipi − L(q, q,˙ t) (16)   i 1 2 2 2 1 2 E = m r˙ + r θ˙ + Iω0 (12) 2 2 asystemof2n + 1 equations is obtained. In this particular case our axes are principal axes ∂H ∂H δL q −p X M ˙i = ; ˙i = ; of inertia, therefore we have δt = 1,andthen, ∂tp ∂qi δ mr2θ˙ i δt ( )=0. (17) And therefore mr2θ˙ = const. ∂L ∂H − In the particular case in which θ˙ =Ωisacon- ∂t = ∂t stant, r has to be a constant (that is,r ˙ has to be If the last equation in (17) is excluded, a system zero) in order to satisfy this equation. We can con- n clude that the body moves in a circular orbit. The of 2 first order equations is obtained, known as energy takes the following expression: canonical Hamilton equations. By substituting H, the first order equations of 1 2 2 1 2 motion, are obtained. E = mr Ω + Iω0 (13) 2 2 It should be noted that the Hamiltonian - tion is developed for holonomous systems and forces Both expressions (11) and (13) have to represent derived from a potential that depends on the po- the same energy. Therefore, comparing the two ex- sition or from generalized potentials. A torque is pressions we can obtain the radius of the orbit: applied to the present system. The result of the

ν0 forces is zero on the centre of mass and, therefore, r = (14) it is meaningless to refer to . Ω ∂V On the other hand, the Hamiltonian function The term − ∂r normally appears in this radial concept does remain meaningful. equation of motion and represents the force derived By using polar coordinate in the movement from a potential. In other words, plane, ˙ ∂V νr =˙r; vθ = rθ (18) mr mrθ˙2 − ¨ = ∂r (27) The Lagrangian is

1 2 2 2 In the present case, this term does not exist. L = m(˙r + r θ˙ ) (19) 2 To complete one’s knowledge of the system’s de- velopment, those relations derived from the con- The generalized moments are straints, have to be resorted to. P mr P mr2θ˙ r = ˙; θ = (20) 1 2 1. E = 2 mν = const. ⇒|ν| = const so that P P     r r θ˙ θ dL dL ˙ = ; = 2 (21) 2. dt = dt + Ω × L m mr XY Z XY Z The Hamiltonian H is introduced using the equa- tion  where the applied external torque H(p, q, t)= q˙ipi − L(q, q,˙ t) (22)   dL i M = dt (28) resulting in XY Z   1 2 1 2 2 H = Prr˙ + Pθθ˙ − mr˙ + mr θ˙ by hypothesis it is known that 2 2 dL dt = 0 and from (1) it is concluded that  XY Z If the generalized velocities are substituted by dL dt = Ω × L the generalized moments, the following is obtained XY Z and it is obtained that Ω = M . P 2 P 2 L H r θ This vector represents the rotation velocity of the = + 2 (23) 2m 2mr frame of axes linked to the cylinder (X, Y , Z) The first pair of Hamilton equations is about the frame of inertial references axes (X, Y , Z). ∂H P ∂H P r r θ˙ θ In polar coordinates, the variable defining the ˙ = ∂P = m ; = ∂P = mr2 r θ rotation about the frame of axes is θ˙ and it is con- The pair of equations is cluded that M ∂H P 2 ∂H θ˙ =Ω= (29) −P˙ − θ −P˙ L r = ∂r = mr3 ; θ = ∂θ =0 substituting in the radial equation of motion, re- The second of these equations shows that the an- sults that J 2 2 gular momentum is conserved. mr¨ = mrθ˙ = mrΩ (30) Pθ = J = const. (24) Moreover, Pθ=const. and at the initial moment The first gives the radical equation of motion, equals P mr2θ˙ mrν J 2 θ = = (31) P˙ mr r = ¨ = 3 (25) mr as ν is constant, it is concluded that r is also 2 constant. Finally it is found that as J = Pθ = mr θ˙, by substituting it results that ν m2r4θ˙2 r mr mν P˙ mrθ˙2 = and ¨ = Ω (32) r = mr3 = (26) Ω Further considerations The Hamiltonian is independent of θ. Thisisan expression of the system’s rotation or, in other words, there is no preferred alignment in the plane. ∂H The equation ∂θ = 0 means that the energy of the system remains invariable if it is turned to a new without changing r, Pr or Pθ.This aspect is clearly proved. Worthy of mention in this case is the fact that the Hamiltonian equations do not provide, in this type of problem, complete information on the ra- dial movement. The reason for this is that the cen- Figure 2. in a magnetic field. tral force is a function of θ which is a coordinate ν that does not appear in the Hamiltonian (it can It is clear that the rotation of 0 causes rotation ν XY ν be ignored) and can only be calculated using the of the 0 component on the plane, 0xy.Fur- constraints. thermore, according to the theory here presented, ν This proposal is valid for any property of the the particle will trace a helix and the 0 compo- Z ν particle that displays vectorial characteristics and nent in the direction of the axis, 0z, will not be is sensitive to an external torque of the type de- affected. L s scribed, without the need to ascertain the real phys- In this case the angular momentum = . B ical nature of that property. The proposal can be If the particle stops inside , the rotational extended to equivalent problems, despite their not velocity of the angular momentum would be ex- involving the presence of internal angular momen- pressed by: tum as a central characteristic. |μ × B | |M | Poisson brackets. The [Pθ,H]is Ω= = (33) |L| |L| ∂Pθ ∂H ∂Pθ ∂H ∂H [Pθ,H]= − = − α s ν ∂θ ∂Pθ ∂Pθ ∂θ ∂θ Obviously, if the formed by and 0 remains constant, the angle γ formed by sxy and ∂H ν ν as ∂θ = 0, it is concluded that [Pθ,H]=0. 0xy will also remain constant. The velocity of axy can be calculated merely by calculating the velocity s CHARGED PARTICLE IN A MAGNETIC of xy. s s θ xy = sin (34) and Let us apply this theory to the specific case of a charged particle with spin s and magnetic momen- |μ||B | sin θ μB Ων0xy =Ωs0xy = = (35) tum μ, moving at a velocity ν0 in a uniform mag- L sin θ L netic field of flux B . As it’s well known, the interaction between B We can conclude from this expression that the and μ makes a torque to act upon the particle. This of the rotation of the particle de- μ situation verifies every condition we have defined in pends on the magnetic momentum , the magnetic B our hypothesis, and so, the behaviour of the particle field in which the particle is immersed and the s must be in agreement with the theoretical model we spin of the particle, and it is independent of the μ B ν have developed above. relative positions of , and 0. Considering the most general of the cases, let us In the case in question, the angular velocity of B suppose that μ forms unknown θ with B the rotation of the particle in is expressed by: and α with ν0. This situation is represented in the e following figure (See Fig. 2). Ω=γ B (36) 2m According to the theory developed above, the particle will draw a circular trajectory. We can determine the radius of the orbit drawn by the elec- tron under the influence of a magnetic field

ν0 2mν0 r = = (37) Ω γeB

In the specific case in which the particle is an , γ has a value of 2 and the radius of the orbit is calculated from the following equation:

ν0 mν0 r = = (38) Ω eB From the conclusions summarized above, we state that the particle is submitted to a affecting the charged particle inside the mag- netic field. To calculate its value, we only need to recall the formula: F = mν0 × Ω (39)

and substitute Ω for its value previously obtained. Figure 3. Airmodel with spinning disc Then we get: e F = mν0 × γ B (40) Basically it consists of a that can rotate, 2m attached on of the flying model, which provides Once again, if we apply this to the case in which the necessary support and . the particle is an electron, the gyromagnetic factor Manoeuvring the flying system, aileron to turn is 2, and hence right and left, elevators to shift upwards and down- wards and rudder, we can obtain the adequate F = eν0 × B (41) to affect this motion. The flying model is controlled by means of a radio The behaviour predicted by the theory here equipment affecting the normal moving elements shown of a charged spinning particle penetrating and the rotation of the disk. ν into a magnetic field with velocity ,matchesthe The fist flight was performed without the disk to one that can be observed in the laboratory, and confirm the normal evolution of the model. the force to which it will be submitted is the well Once the normal behaviour of the model in this known . conditions was confirmed, the disk was attached on . top of the model. EXPERIMENTS CARRIED OUT . With the disk still the system behaves in a sim- . ilar way that it did without the disk, although the Airmodel with spinning disc stability was slightly affected. The airmodel is provided of a strong angular mo- With the disk rotation pointer clock way when we mentum by mean of a high speed spinning disc at- actuate the ailerons to the left, instead of drawing tached to it by exerting different torques on it, with a horizontal , it actually goes up even the built in flight systems, we achieve modifications to the of performing a loop. in its flight trajectory. Actuating the elevators upwards causes the The remote control flying model was designed model to turn right, when we actuate the elevators specifically to verify this theory (See Fig. 3). downwards the model turns to the left. . . In agreement with this theory, the is Magnetic spinning top affected by the force . Withthisexperimentitisprovedthatthedirec- F = mν × Ω (44) tion way of the precession can be manipulated in accordance with the torque applied. The torque will be For the this experiment we have a magnetized , a aluminium and a magnet. M = r × mν × Ω= L × Ω (45) We attach the rod to the cone. We have build up a magnetized spinning top (See Fig. 7). (It can be proved that the associative property of the vectorial product is satisfied in this case) Substituting (45) in (42) we obtain:

dL L × dt = Ω (46)

This one is the equation of the motion of a vector with constant module |L |, that precesses around the axis defined by Ω, with angular velocity |Ω |.

Larmor Effect (Samples under the influence of mag- netic fields) In agreement with this theory, an electron em- Figure 7. Magnetic spinning top. bedded inside an in the presence of a mag- B netic field , is submitted to the central force We observe the behaviour of the magnetized spinning top in the presence of a magnetic field. F = mν × Ω (47) The interaction between both fields produces a torque which affects the spinning top making him which adopts the expression to draw a circular trajectory. qB q The spinning top will not collapse on the mag- F mν × mν × ν × B = Ω= mc = c (48) neto as long as it has an angular momentum.

and is known as the Lorentz Force. EXPERIMENT SUGGESTED TO TEST As the most general expression, we can write: THIS THEORY: INTERACTION OF A HOMOGENEOUS dL AND A PERPENDICULAR GYRATING = r × mν × Ω= L × Ω (49) dt MAGNETIC FIELD, WITH A PARTICLE WITH SPIN AND (It can be proved that the associative property of the vectorial product is satisfied in this case) Formulation of the problem This is the equation for the motion of L ,which rotates around the vector B with an angular veloc- Theparticlewithspin,S, moves with velocity ν.It ity: B q first passes into a homogeneous magnetic field , − B inducing the magnetic moment, μ, of the particle, Ω= mc (50) polarized in the direction of the field. It later enters The general expression of the central force, which a region in which a homogeneous field, B ,anda includes the Lorentz force as a particular case, al- gyrating field, H0, with rotation , φ˙,are lows us to explain the Larmor effect and its mean- superposed. Fields B and B are parallel and in ing in an easy and accurate way within the range fact could even be the same field, although their of a wider and more important phenomenon. function is different in each region. Reference frame (See Fig. 8). It is proved that ψ˙ = ωZ . The proof is easy, for if XYZ Fixed system in the particle but which the Z axis is fixed, i.e. both φ and θ are constant, does not rotate with this. the about the Z axis are rotations of the B is on the Z axis. X and Y axes in the XY plane. Hence, the Euler H0 is on the XY plane. angle that reports the rotation is ψ.Itcanthenbe XY Z Fixed frame in the particle which pre- concluded that cesses with this. ψ˙ = ωZ (54) S and μ are on the Z axis. By agreement, but without loss of generality, let ν is on the Y axis. us consider that the particle has an angular mo- X0Y0Z0 Reference frame from which observa- mentum, S, that is constant in module. tions are made. B is on the Z0 axis. H0 is on the X0Y0 plane. Frame of axes linked to the solid In the definition of the frame of axes linked to the solid, we can either choose:

a) The frame of axes strictly accompanies the particle in the rotation which, in classical terms, gives its internal angular momentum.

b) The frame of axes linked to the solid is de- fined by a parallel axis to the internal angu- lar momentum of the particle, and the other two are perpendicular to each other and are in a plane which is perpendicular to the angu- lar momentum. Internal angular momentum is understood to be a quality of the particle whose mathematical characteristics coincide with those of the angular momentum, without Figure 8. Reference frames. considering the physical nature of this quality. In other words, no hypothesis is formed as to The rotation velocity, Ω, of the frame of axes whether or not the angular momentum implies linked to the solid, XY Z, will be expressed within rotations of the particle under study. this system as: For the purposes of our problem, definition (b) Ω=Ω1i +Ω2j +Ω3k (51) has been used. However, it should be remembered that our par- This rotation velocity, Ω, can also be expressed ticle can reach a rotation velocity, ω,asaresultof bwithy using respect Euler to angles the as: axes, XYZ the interactions to which it can be subjected. by using . of the interactions Ω1 = φ˙ sin θ sin ψ + θ˙ cos ψ Ω2 = φ˙ sin θ cossin ψ − θ˙ cossin ψ (52) Given the nature of our problem, the particle pen- etrates a homogeneous magnetic field, B , inducing Ω3 = φ˙ cos θ + ψ˙ the magnetic moment, μ, of the particle, polarized Our particle can also rotate with respect to the in the direction of the field. frame of axes, XY Z, and the most general ex- Hence, when the particle penetrates the magnetic field, B (parallel to B , and can even coincide with pression to describe this angular velocity is: it) and H0, the magnetic moment, μ, is only (ini- ω = ωX i + ωY j + ωZ k (53) tially) sensitive to the gyrating magnetic field, H0. This, consequently, will be the first interaction that we analyze.

Interaction with the magnetic field, H0 The interaction of the gyrating magnetic field, H0, with the magnetic moment, μ,doesnotmodifythe energy of the system, as both are permanently per- pendicular. The energy of the interaction is expressed

π EH = H0 · μ = H0μ cos = 0 (55) 2 Figure 9. Interaction with the magnetic field, H0.

The interaction between H0 and μ,isequivalent μH to the of a torque, ΓH , on the particle, where θ 0 = S φ˙ = 0 (59) ΓH = μ × H0 (56) Ω3 =0

ΓH is perpendicular to the X and Z axes, and in other words, the rotation of the frame linked to XY Z θ˙ is located on the Y axis. the solid on ,Ω1, is constant and equal to , where θ˙ is the rotation of the frame of axes linked By using the previously obtained Euler equations to the solid with regard to XYZ expressed in the (9) and applying them to our problem: Euler angles. We have obtained the rotation velocity, θ˙, but we M3 .. r t ω˙ Z = =0=ψ˙ have to calculate the rotation radio ( )ifwewant I3 to know the trajectory. To do this we will use the ΓH μH0 following between differential operators, Ω1 = S = S (57) ∗ M d d 1 = + Ω × , Ω2 = S =0 dt dt .. where ψ˙ M d so that = 0 in all cases,. given that 3 = 0 at all dt is the derivative in the inertial frame, in our case . We can avoid ψ by making it equalconstant to in zero all X0Y0Z0, d∗ incases, all caseand for (See convenienc Fig. 9). e y = 0 (See Fig. 9). dt is the derivative in the non-inertial frame, in our case XY Z, The Euler equations depending on the Euler an- gles (52), are simplified with the result that Ω is the angular rotation velocity of the frame of axes linked to the solid in relation to the inertial μH frame of axes. θ˙ 0 r Ω1 = = S We will apply this equality to , a vector of posi- tion of one frame with respect to another, and for φ˙ θ Ω2 = sin = 0 (58) convenience we will express all the vectors in the Ω3 = φ˙ cos θ X Y Z frame. d∗r dr − × r from which it is deduced that dt = dt Ω (60) If we represent the equation in components, we Interaction with the magnetic field B have At the moment the circular trajectory is com- menced, θ is no longer null and the interaction with r˙X + rZ Ω2 − rY Ω3 = νX 0 the field, B , begins (See Fig. 11). r r − r ν ˙Y + X Ω3 Z Ω1 = Y 0 (61)

r r − r ν ˙Z + Y Ω1 X Ω2 = Z 0 Γ=μ × B (63) μB sin θ Ω2 = - (64) Taking into account the initial conditions (See S Fig. 10), Ω2 =0andΩ3 =0.

Figure 11. Interaction with the magnetic field.

Figure 10. Initial conditions. M3 continues to be null and, therefor, so do ψ˙ and ψ, which do not vary, but Ω2 is now different The interaction does not modify the energy of the from zero. system, as explained beforehand, so the module of θ˙ ν has to remain constant. Ω1 = μB θ φ˙ θ - sin Ω2 = sin = S (65) |ν| = ν = constant. Ω3 = φ˙ cos θ and, νX =0,νY = ν, νZ =0. 0 0 0 We finally obtain, As the energy of the system cannot vary,r ˙X = r˙Y =˙rZ = 0. If they differed from zero tangen- μH θ˙ 0 tial would be involved and therefore = S a variation in the total energy of the system. μB φ˙ - These conditions simplify the equations of our = S (66) system (61) and we obtain μB cos θ Ω3 = S ν rZ = − (62) Ω1 These are the expressions of the rotation of the frame of axes linked to the solid in the that This is to say, the particle and the frame of axes defines the trihedral linked to the solid. linked to the particle describe a circular trajectory We can also express this same rotation velocity of with respect to the observation frame. The move- the frame of axes linked to the solid with respect to ment plane is perpendicular to H0. the reference frame, by means of the Euler angles, but in the base defining the trihedral of reference The behaviour of the particle indicates that it is frame. subjected to two central forces F mν × mν × θ˙i ωX = θ˙ cos φ + ψ˙ sin θ sin φ 1 = Ω1 = (70) ω θ˙ φ − ψ˙ θ φ Y = sin sin cos (67) F2 = mν × φ˙k (71) ω ψ θ φ˙ Z = cos + From an energy point of view, we can confirm that the magnetic field, H0,doesnotmodifythe As ψ˙ = ψ =0,weobtain energy of the system, and that the magnetic field, B , cyclically modifiesthe the potencial potential and energy kinetic of energy the ωX = θ˙ cos φ particle,of the particle, since since ωY = θ˙ sin φ (68) EB B · μ θ ωZ = φ˙ = cos (72) These expressions clearly reflect the behaviour of μH θ t 0 t the frame of axes linked to the solid and of the solid, with ( )= S with respect to the reference axes. but with its unvaried. Visualization of the movement is assisted by con- sidering the following specific situations φ˙ =0 Additional considerations i) For φ =0 The reader will have already noted the ωX = θ˙ between this behaviour and the Larmor effect. ωY =0 It is know that this effect is produced by applying a magnetic field, B , to a particle of charge, q,mov- ωZ =0 ing in an orbit around a fixed specific charge, q . The result is a precession of the trajectory around the direction of the applied magnetic field, with a X qB We would have a rotation about the axis. precession velocity, ωL = 2m ,knowasLarmorfre- quency, with the proviso that the cyclotronic fre- j) For φ = π 2 quency is directly obtained by this procedure. This statement can be expressed by saying that ωX =0 ω θ˙ the angular momentum vector of the particle with Y = respect to the rotation axis, referred to as orbital ωZ =0 angular momentum, precesses about the direction of the magnetic field, B . From our procedure, it is particularly easy to We would have a rotation about the Y axis. reach the same conclusions by considering that The rotation velocity in the XY plane is dL = ΓB = r × F dt ωXY = θ˙(cos θi +sinθj) (73) (69) L = r × mν |ωXY | = θ˙ By substituting our development, F = mν × Ω, Therefore, the total movement is a circular tra- in (73) we obtain jectory of radius, rZ , perpendicular to the field, H0, which in turn spins about the field, B,with dL = r × (mν × Ω) velocity φ˙ (classically know as Larmor rotation fre- dt quency). This explains why the frequency of the (74) H oscillating field, 0, has to coincide with the Lar- =(r × mν) × Ω= L × Ω mor frequency. Figure 12. Rabi’s experiment.

Figure 13. Experiment suggested.

(the triple vector product does not generally Experimental conditions fulfill the associative property, but it can be de- mostrated in our.) This experiment can be carried out by emulating Equation (74) describes the angular momentum the rotating magnetic field using a radio frequency precession about the magnetic field with a preces- magnetic field, just as Rabi did in his experiments. sion rate Ω. Rabi’s experiment (see References (2)) consists The Larmor effect results from subjecting the of a collimated particle beam that crosses an inho- particle to the Lorentz central force, because it is mogeneous magnetic field. It later passes through a within B. From the following, the Lorentz force is region where a homogeneous and a radio frequency equivalent to our force, given that magnetic field are superposed, and finally passes μB e through an inhomogeneous field that refocuses the F mν mν mν B eνB beam towards the detector. = Ω= S = m = (75) The inhomogeneous fields separate the beam into Which demostrates that both movements are different beams according to their magnetic mo- equivalent. ment (the dependence this on the spin) as in an experiment of Stern-Gerlach. When leaving the CONCLUSIONS first field, these beams later pass into the second inhomogeneous field, which refocuses the beams to When a system with internal angular momentum, the detector. By adjusting the second inhomoge- moving at constant velocity with respect to a ref- neous magnetic field we will obtain refocusing con- erence frame, is subjected to an interaction of the ditions for a beam or group of beams. type under study, i.e., a torque perpendicular to the In the central part of the experimental arrange- internal angular momentum vector, it will begin to ment, the homogeneous magnetic field is super- trace a circular path of radius posed on the radio frequency field. In this region, ν r = the spin and magnetic moment are deflected with Ω respect to the constant field when the oscillating M field frequency approaches the where Ω = L frequency. This process is know as nuclear mag- Its behaviour is equivalent to that produced by netic . subjecting the cylinder to a central force After deflection, the atom is on another level F mν × which will not fulfill the refocusing condition and = Ω a decrease in the beam will be observed As it can be seen in figure 14, the trajectory in the detector. This procedure is used to study which particle follows is the trajectory II, while the nuclear spin, nuclear magnetic moments and hy- trajectory I is the intuitive one, but as it has been perfine structures. However, the behaviour of the shown in this paper it is not the real behaviour of in Rabi’s experiment is theoretically jus- the particle. tified in our exposition and it is therefore unnec- essary to use inhomogeneous magnetic fields (as is ACKNOWLEDGMENTS required in Rabi’s experiments). Our experiment only requires the presence of a Through the development of this paper I have had homogeneous magnetic field to induce the magnetic the privilege of maintaining some discussions about moment, μ, of the particle, polarized in the direc- its content with Dr. Jos´eL.S´anchez G´omez. tion of the field. It will also like to thank Miguel Morales Furi´o In such a way that: for his support and help to the development of this o article. 1 Only those particles whose Larmor frequency, In the same way, I will like to express my grat- φ˙ , coincides with the radio frequency will itude to Juan Silva Trigo, Jes´us Abell´an, Berto abandon the rectilinear trajectory. Gonz´alez Carrera and to all the other people who

o has worked in the design and realization of all the 2 The trajectory described by these particles experiments describe in this article, whose enumer- should coincide with that of the theoretical so- ation would be too extensive. lution given in this article while they remain I would also like to express that the responsabil- within the region in which the two fields co- ity for any affirmation here introduced lies exclu- exist. According to this solution, the parti- sively with the author. cle abandons the rectilinear path and the XY plane.

3o When the particle abandons the region of co- existing fields, it will follow a rectilinear move- ment but will not reach the detector.

4o This procedure facilitates selection of the dif- ferent angular moments because of their de- pendence on the Larmor precession. Figure 14. Classical trajectory and trajectory predicted in this article.

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